# ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO

**Solution:**

We know according to the basic proportionality theorem, if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.

Consider the trapezium ABCD as shown below.

In trapezium ABCD,

AB || CD

Also, AC and BD intersect at ‘O’

Construct XY parallel to AB and CD (XY || AB, XY || CD) through ‘O’

In ΔABC

OY || AB (construction)

According to theorem 6.1 (Basic Proportionality Theorem)

BY/CY = AO/OC................. (1)

In ΔBCD

OY || CD (construction)

According to theorem 6.1 (Basic Proportionality Theorem)

BY/CY = OB/OD................. (2)

From equations (1) and (2)

OA/OC = OB/OD

⇒ OA/OB = OC/OD

Hence proved.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**Video Solution:**

## ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO

Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 9

**Summary:**

Hence it is proved that AO/BO = CO/DO if ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.

**☛ Related Questions:**

- In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)
- E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF || QR : (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
- In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
- In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.

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